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Misc 26 - Show that 1 x 22 + 2 x 32 - Chapter 9 Class 11 Series - Finding sum from nth number

  1. Chapter 9 Class 11 Sequences and Series
  2. Serial order wise
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Misc 26 Show that (1 × 22 + 2 × 32 + … + n × (n + 1)2)/(12 × 2 + 22 × 3 + … + n2 × (n + 1)) = (3n + 5)/(3n + 1) Taking L.H.S (1 × 22 + 2 × 32 + … + n × (n + 1)2)/(12 × 2 + 22 × 3 + … + n2 × (n + 1)) We solve denominator & numerator separately Solving numerator Let numerator be S1 = 1 × 22 + 2 × 32 + … + n × (n + 1)2 nth term is n × (n + 1)2 Let an = n(n + 1)2 = n(n2 + 1 + 2n) = n3 + n + 2n2 Now finding S1 = ((𝑛(𝑛 + 1))/2)^2 + 2(( 𝑛(𝑛+1)(2𝑛+1))/6) + n(n+1)/2 = (𝑛(𝑛 + 1))/2 (n(n+1)/2 " + " (2(2𝑛+1))/3 " + 1" ) = (𝑛(𝑛 + 1))/2 (( 3𝑛(𝑛+1) + 2 × 2(2𝑛+1)+ 6)/6) = (n(n + 1))/(2 × 6)[3n(n + 1) + 4(2n + 1) + 6] = (n(n + 1))/12[3n2 + 3n + 8n + 4 + 6] = (𝑛(𝑛 + 1))/12[3n2 + 11n + 10] = (𝑛(𝑛 + 1))/12[3n2 + 5n + 6n + 10] = (𝑛(𝑛 + 1))/12[n(3n + 5) + 2(3n + 5)] = (𝑛(𝑛 + 1))/12[(n + 2)(3n + 5)] Thus, S1 = (𝑛(𝑛 + 1))/12[(n + 2)(3n + 5)] Now solving denominator Let denominator be S2 = 12 × 2 + 22 × 3 + … + n2 × (n + 1) nth term is n2(n + 1) Let bn = n2(n + 1) bn = n3 + n2 Now, calculating S2 = ((𝑛(𝑛 + 1))/2)^2 + (( 𝑛(𝑛+1)(2𝑛+1))/6) = (𝑛(𝑛 + 1))/2 (n(n+1)/2 " + " ((2𝑛+1))/3) = (𝑛(𝑛 + 1))/2 (n(n+1)/2 " + " ((2𝑛+1))/3) = (𝑛(𝑛 + 1))/2 (( 3𝑛(𝑛+1) + 2 (2𝑛+1))/6) = (n(n + 1))/(2 × 6) (3n(n + 1) + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 4n + 2) = (n(n+1))/12 (3n2 + 7n +2) = (n(n+1))/12 (3n2 + 6n + n +2) = (n(n+1))/12 (3n(n + 2) + 1(n +2)) = (n(n+1)(n+2)(3n+1))/12 Thus, S2 = (n(n+1)(n+2)(3n+1))/12 Now, Taking L.H.S (1 × 22 + 2 × 32 + … + n × (n + 1)2)/(12 × 2 + 22 × 3 + … + n2 × (n + 1)) = 𝑆1/𝑆2 = ((n(n+1)(n+2)(3n+5))/12)/((n(n+1)(n+2)(3n+1))/12) = (n(n+1)(n+2)(3n+5))/12 × 12/(n(n+1)(n+2)(3n+1)) = (n(n+1)(n+2)(3n+5))/(n(n+1)(n+2)(3n+1)) = ((3n+5))/((3n+1)) = R.H.S Hence L.H.S = R.H.S Hence proved.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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